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Chapter 15: Problem 41

A \(3.00-\mathrm{m}\) -long string, fixed at both ends, has a mass of \(6.00 \mathrm{~g}\). If you want to set up a standing wave in this string having a frequency of \(300 . \mathrm{Hz}\) and three antinodes, what tension should you put the string under?

Short Answer

Expert verified
To find the tension required in the string, follow these steps: 1. Calculate the linear mass density: μ = 0.00200 kg/m. 2. Calculate the period: T = 0.00333 s. 3. Determine the wavelength: λ = 2.00 m. 4. Calculate the wave velocity: v = 600 m/s. 5. Calculate the tension: T = 720 N. The tension required is 720 N.

Step by step solution

01

Determine the linear mass density and period

First, we'll calculate the linear mass density \(μ\) by dividing the mass of the string by its length: $$ μ = \frac{m}{L} = \frac{6.00\,\text{g}}{3.00\,\text{m}} = 2.00\,\text{g/m} = 0.00200\,\text{kg/m} $$ Next, we'll find the period \(T\) using the given frequency \(f\): $$ T = \frac{1}{f} = \frac{1}{300\,\text{Hz}} = 0.00333\,\text{s} $$
02

Find the wavelength

Since we have 3 antinodes in the string, it means we have \(1.5\) complete wavelengths over the length of the string. Therefore, we can find the wavelength \(λ\) by dividing the length of the string by \(1.5\): $$ λ = \frac{L}{1.5} = \frac{3.00\,\text{m}}{1.5} = 2.00\,\text{m} $$
03

Calculate the wave velocity

Now we have the wavelength \(λ\) and the period \(T\), we can find the wave velocity \(v\) using the wave equation: $$ v = \frac{λ}{T} = \frac{2.00\,\text{m}}{0.00333\,\text{s}} = 600\,\text{m/s} $$
04

Calculate the tension

Finally, we can use the formula \(v = \sqrt{\frac{T}{μ}}\) to determine the required tension in the string to create the desired standing wave: $$ T = μv^2 = (0.00200\,\text{kg/m})(600\,\text{m/s})^2 = 720\,\text{N} $$ The tension required to create the desired standing wave with a frequency of \(300\,\text{Hz}\) and three antinodes in the string is \(720\,\text{N}\).

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Most popular questions from this chapter

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

Students in a lab produce standing waves on stretched strings connected to vibration generators. One such wave is described by the wave function \(y(x, t)=(2.00 \mathrm{~cm}) \sin \left[\left(20.0 \mathrm{~m}^{-1}\right) x\right] \cos \left[\left(150 . \mathrm{s}^{-1}\right) t\right],\) where \(y\) is the transverse displacement of the string, \(x\) is the position along the string, and \(t\) is time. Rewrite this wave function in the form for a right- moving and a left-moving wave: \(y(x, t)=\) \(f(x-v t)+g(x+v t)\); that is, find the functions \(f\) and \(g\) and the speed, \(v\)

A string is \(35.0 \mathrm{~cm}\) long and has a mass per unit length of \(5.51 \cdot 10^{-4} \mathrm{~kg} / \mathrm{m}\). What tension must be applied to the string so that it vibrates at the fundamental frequency of \(660 \mathrm{Hz?}\)

Noise results from the superposition of a very large number of sound waves of various frequencies (usually in a continuous spectrum), amplitudes, and phases. Can interference arise with noise produced by two sources?

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